Question

A block is attached to a spring and it oscillates with natural frequency \(f_1\). If the spring is cut into two half and only one of the half spring is connected to the block then the frequency becomes \(f_2\). Find \( \frac{f_2}{f_1} \).

• A. \( \sqrt{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \(2\)
• D. \( \frac{1}{2} \)

Answer 1

The Correct Option is A

Concept: The frequency of oscillation of a spring-block system is \[ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \] where \(k\) = spring constant \(m\) = mass attached to the spring When a spring is cut into equal halves, the spring constant of each half becomes twice the original.
Step 1:Write the original frequency of the spring-block system. \[ f_1 = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \]
Step 2:Determine the new spring constant after cutting the spring. When the spring is cut into two equal parts, \[ k' = 2k \]
Step 3:Calculate the new frequency. \[ f_2 = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} \] \[ f_2 = \sqrt{2}\left(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\right) \] \[ f_2 = \sqrt{2}\, f_1 \]
Step 4:Find the required ratio. \[ \frac{f_2}{f_1} = \sqrt{2} \] \[ \boxed{\frac{f_2}{f_1} = \sqrt{2}} \]